Question

Resources A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point) the homeowner will have a better chance of moving the rock. The homeowner places the fulcrum a distance 0.266 m from the rock, which has a mass of 325 kg, and fits one end of the rod under the rock's center of weight If the homeowner can apply a maximum force of 663 N at the other end of the rod. what is the minimum focal length of the rod required to move the rock? Assume that the rod is massless and nearly horizontal so that the weight of the rock and homeowner's force are both essentially vertical No P2

Answer 1

mass of rock, m= 325 kg distance of Julcrum from rock, d= 0.766m In order to move the rock, The Torque produced by Force Opleied by that man should be atleast equal to the Torque produced by weight of rock, about Sulerum. so taking moment about Lulerum [upLength on 7 S l Right of fulcrum (mg) x d = F XLR L' Lo= mg d 325X 9.81X 0.266 663 L = 1.279 m . R so total length of rod L = d+LR = 0.266 +1.779 T[= 1.545 m o Answer Answer